\(\displaystyle \begin{array}{{>{\displaystyle}l}} x-4+\sqrt{16-8x+x^{2}}\\ \rightarrow x-4-\sqrt{( 4-x)^{2}}\\ \rightarrow x-4-|4-x|\\ Với\ \left[ \begin{array}{l l} 4-x\geqslant 0\ \rightarrow \ x-4-4+x=2x-8 & \\ 4-x< 0\ \rightarrow \ x-4+4-x=0\ & \end{array} \right. \end{array}\)

Ta có : tg60=m-1

\({\sqrt{3}=m-1} \) \(->m=\sqrt{3} +1\)

\(tan120=3-2m <=> -\sqrt{3}=3-2m \)

m=\(\frac{3+\sqrt{3}}{2}\)